Simplify and expand the following expression: $ \dfrac{3}{2y + 16}+ \dfrac{2}{3y + 15}+ \dfrac{3y}{y^2 + 13y + 40} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the first term: $ \dfrac{3}{2y + 16} = \dfrac{3}{2(y + 8)}$ We can factor a $3$ out of denominator in the second term: $ \dfrac{2}{3y + 15} = \dfrac{2}{3(y + 5)}$ We can factor the quadratic in the third term: $ \dfrac{3y}{y^2 + 13y + 40} = \dfrac{3y}{(y + 8)(y + 5)}$ Now we have: $ \dfrac{3}{2(y + 8)}+ \dfrac{2}{3(y + 5)}+ \dfrac{3y}{(y + 8)(y + 5)} $ The least common multiple of the denominators is: $ 6(y + 8)(y + 5)$ In order to get the first term over $6(y + 8)(y + 5)$ , multiply by $\dfrac{3(y + 5)}{3(y + 5)}$ $ \dfrac{3}{2(y + 8)} \times \dfrac{3(y + 5)}{3(y + 5)} = \dfrac{9(y + 5)}{6(y + 8)(y + 5)} $ In order to get the second term over $6(y + 8)(y + 5)$ , multiply by $\dfrac{2(y + 8)}{2(y + 8)}$ $ \dfrac{2}{3(y + 5)} \times \dfrac{2(y + 8)}{2(y + 8)} = \dfrac{4(y + 8)}{6(y + 8)(y + 5)} $ In order to get the third term over $6(y + 8)(y + 5)$ , multiply by $\dfrac{6}{6}$ $ \dfrac{3y}{(y + 8)(y + 5)} \times \dfrac{6}{6} = \dfrac{18y}{6(y + 8)(y + 5)} $ Now we have: $ \dfrac{9(y + 5)}{6(y + 8)(y + 5)} + \dfrac{4(y + 8)}{6(y + 8)(y + 5)} + \dfrac{18y}{6(y + 8)(y + 5)} $ $ = \dfrac{ 9(y + 5) + 4(y + 8) + 18y} {6(y + 8)(y + 5)} $ Expand: $ = \dfrac{9y + 45 + 4y + 32 + 18y}{6y^2 + 78y + 240} $ $ = \dfrac{31y + 77}{6y^2 + 78y + 240}$